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3 years ago

Simplest form of 14/63

Mathematics

2 answers:

docker41 [41]3 years ago

7 0

Answer:

2/9

Step-by-step explanation:

kupik [55]3 years ago

5 0

Answer:

2/9

Step-by-step explanation:

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A typical sugar cube has an edge length of 1.00 cm. Imagine you have a cubical box that contains a mole of sugar cubes, perfectl

alisha [4.7K]

Answer:

844368.7 m

Step-by-step explanation:

We are given that a typical sugar cube has an edge length of 1.00 cm.

We have to find the edge length of the box in meters.

Edge length of sugar cube=1 cm = $10^{-2}m$ ( $1 cm=\frac{1}{100}m$ )

Volume of a sugar cube= $side\times side\time side=(10^{-2})^3=10^{-6}m^3$

1 mole of sugar= $6.02\times 10^{23}units$

Volume of 1 unit= $10^{-6} m^3$

Volume of $6.02\times 10^{23}$ units= $6.02\times10^{23}\times 10^{-6}=6.02\times 10^{17} m^3$

Volume of box=1 mole of sugar= $6.02\times 10^{17} m^3$

Edge length of box= $\sqrt[3]{6.02\times 10^{17}}=844368.7 m$

6 0

3 years ago

Pls help i only have 2 minutes!!!!!!!!!!!!!!

umka21 [38]

Answer:

40 + ( -63) is less

Step-by-step explanation:

.... .....

4 0

2 years ago

Read 2 more answers

Find the arc length of the partial circle.

Kay [80]

Answer:

12.42 units

Step by step explanation:

Given,

length of the radius = 2 units

Therefore, arc length of the complete circle = 2 × 3.14 × 2 units = 12.56 units

Therefore, arc length of the partial circle = 3/4 × 12.56 units

= 12.42 units

3 0

3 years ago

Find the exact value of each trigonometric function for the given angle θ.

Kay [80]

Answer:

$\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=\dfrac{2}{\sqrt{3}}.$

Step-by-step explanation:

The given angle is 240 degrees.

We need to find the exact value of each trigonometric function for the given angle θ.

Since $\theta=240$ , it means θ lies in 3rd quadrant. In 3d quadrant only tan and cot are positive.

$\sin (240^\circ)=\sin (180^\circ+60^\circ)=-\sin (60^\circ)=-\dfrac{\sqrt{3}}{2}$

$\cos (240^\circ)=\cos (180^\circ+60^\circ)=-\cos (60^\circ)=-\dfrac{1}{2}$

$\tan (240^\circ)=\tan (180^\circ+60^\circ)=\tan (60^\circ)=\sqrt{3}$

$\cot (240^\circ)=\cot (180^\circ+60^\circ)=\cot (60^\circ)=\dfrac{1}{\sqrt{3}}$

$\sec (240^\circ)=\sec (180^\circ+60^\circ)=-\sec (60^\circ)=-2$

$\csc (240^\circ)=\csc (180^\circ+60^\circ)=-\csc (60^\circ)=-\dfrac{2}{\sqrt{3}}$

Therefore, $\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=-\dfrac{2}{\sqrt{3}}.$

8 0

2 years ago

Can any of you help me with this

RUDIKE [14]

If f(x)=2x^2+5√(x+2) then:

f(2)=2(2^2)+5√(2+2)

f(2)=2(4)+5√4

f(2)=8+5(2)

f(2)=8+10

f(2)=18

We assume the square root is positive unless otherwise indicated because this is a function and you can have only one y value for each x value to be a function.

6 0

3 years ago